\(x^2 = 2\) has No Solution in \(\mathbb{Q}\)

Proof (by constradiction). Assume that \(x^2 = 2\) has a solution in \(\mathbb{Q}\), i.e. let \(x = \frac{p}{q}\), where \(p, q \in \mathbb{Z}\), \(q \neq 0\), and \(p\) and \(q\) are in lowest terms, i.e. they have no common factors.

So, \(\left(\frac{p}{q}\right)^2 = 2\), hence \(p^2 = 2q^2\). Then \(p^2\) is even (divisible by 2). Then \(p\) is even (because if \(p\) were odd, \(p^2\) would be odd).